Solve sqrt{x^2+3x}+sqrt{x^2-1}=2

$$\sqrt{ { x }^{ 2 } +3x } + \sqrt{ { x }^{ 2 } -1 } =2$$

Solve for x

$$x=1$$

Solution Steps

Subtract \sqrt{x^{2}-1} from both sides of the equation.

$$\sqrt{x^{2}+3x}=2-\sqrt{x^{2}-1}$$

Square both sides of the equation.

$$\left(\sqrt{x^{2}+3x}\right)^{2}=\left(2-\sqrt{x^{2}-1}\right)^{2}$$

Calculate \sqrt{x^{2}+3x} to the power of 2 and get x^{2}+3x.

$$x^{2}+3x=\left(2-\sqrt{x^{2}-1}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2-\sqrt{x^{2}-1}\right)^{2}.

$$x^{2}+3x=4-4\sqrt{x^{2}-1}+\left(\sqrt{x^{2}-1}\right)^{2}$$

Calculate \sqrt{x^{2}-1} to the power of 2 and get x^{2}-1.

$$x^{2}+3x=4-4\sqrt{x^{2}-1}+x^{2}-1$$

Subtract 1 from 4 to get 3.

$$x^{2}+3x=3-4\sqrt{x^{2}-1}+x^{2}$$

Subtract 3+x^{2} from both sides of the equation.

$$x^{2}+3x-\left(3+x^{2}\right)=-4\sqrt{x^{2}-1}$$

To find the opposite of 3+x^{2}, find the opposite of each term.

$$x^{2}+3x-3-x^{2}=-4\sqrt{x^{2}-1}$$

Combine x^{2} and -x^{2} to get 0.

$$3x-3=-4\sqrt{x^{2}-1}$$

Square both sides of the equation.

$$\left(3x-3\right)^{2}=\left(-4\sqrt{x^{2}-1}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-3\right)^{2}.

$$9x^{2}-18x+9=\left(-4\sqrt{x^{2}-1}\right)^{2}$$

Expand \left(-4\sqrt{x^{2}-1}\right)^{2}.

$$9x^{2}-18x+9=\left(-4\right)^{2}\left(\sqrt{x^{2}-1}\right)^{2}$$

Calculate -4 to the power of 2 and get 16.

$$9x^{2}-18x+9=16\left(\sqrt{x^{2}-1}\right)^{2}$$

Calculate \sqrt{x^{2}-1} to the power of 2 and get x^{2}-1.

$$9x^{2}-18x+9=16\left(x^{2}-1\right)$$

Use the distributive property to multiply 16 by x^{2}-1.

$$9x^{2}-18x+9=16x^{2}-16$$

Subtract 16x^{2} from both sides.

$$9x^{2}-18x+9-16x^{2}=-16$$

Combine 9x^{2} and -16x^{2} to get -7x^{2}.

$$-7x^{2}-18x+9=-16$$

Add 16 to both sides.

$$-7x^{2}-18x+9+16=0$$

Add 9 and 16 to get 25.

$$-7x^{2}-18x+25=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -7x^{2}+ax+bx+25. To find a and b, set up a system to be solved.

$$a+b=-18 ab=-7\times 25=-175$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -175.

$$1,-175 5,-35 7,-25$$

Calculate the sum for each pair.

$$1-175=-174 5-35=-30 7-25=-18$$

The solution is the pair that gives sum -18.

$$a=7 b=-25$$

Rewrite -7x^{2}-18x+25 as \left(-7x^{2}+7x\right)+\left(-25x+25\right).

$$\left(-7x^{2}+7x\right)+\left(-25x+25\right)$$

Factor out 7x in the first and 25 in the second group.

$$7x\left(-x+1\right)+25\left(-x+1\right)$$

Factor out common term -x+1 by using distributive property.

$$\left(-x+1\right)\left(7x+25\right)$$

To find equation solutions, solve -x+1=0 and 7x+25=0.

$$x=1 x=-\frac{25}{7}$$

Substitute 1 for x in the equation \sqrt{x^{2}+3x}+\sqrt{x^{2}-1}=2.

$$\sqrt{1^{2}+3\times 1}+\sqrt{1^{2}-1}=2$$

Simplify. The value x=1 satisfies the equation.

$$2=2$$

Substitute -\frac{25}{7} for x in the equation \sqrt{x^{2}+3x}+\sqrt{x^{2}-1}=2.

$$\sqrt{\left(-\frac{25}{7}\right)^{2}+3\left(-\frac{25}{7}\right)}+\sqrt{\left(-\frac{25}{7}\right)^{2}-1}=2$$

Simplify. The value x=-\frac{25}{7} does not satisfy the equation.

$$\frac{34}{7}=2$$

Substitute 1 for x in the equation \sqrt{x^{2}+3x}+\sqrt{x^{2}-1}=2.

$$\sqrt{1^{2}+3\times 1}+\sqrt{1^{2}-1}=2$$

Simplify. The value x=1 satisfies the equation.

$$2=2$$

Equation \sqrt{x^{2}+3x}=-\sqrt{x^{2}-1}+2 has a unique solution.

$$x=1$$